CPSC 221: Algorithms and Data Structures

Lecture #6Balancing Act

Steve Wolfman

2014W1

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Today’s Outline

• Addressing one of our problems• Single and Double Rotations• AVL Tree Implementation

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Beauty is Only (log n) Deep

• Binary Search Trees are fast if they’re shallow:– perfectly complete– perfectly complete except the one level fringe (like a heap)– anything else?

What matters here?Problems occur when onesubtree is much taller than the other! 3

Balance

• Balance– height(left subtree) - height(right subtree)– zero everywhere perfectly balanced– small everywhere balanced enough

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Balance between -1 and 1 everywhere maximum height of ~1.44 lg n 4

AVL Tree Dictionary Data Structure

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• Binary search tree properties– binary tree invariant– search tree invariant

• Balance invariant– balance of every node is:

-1 b 1– result:

• depth is (log n)

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5Note that (statically… ignoring how it updates) an AVL tree is a BST.

Testing the Balance Property

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NULLs have height -1

6FIRST calculate heightsTHEN calculate balances

An AVL Tree

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data

height

children

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Today’s Outline

• Addressing one of our problems• Single and Double Rotations• AVL Tree Implementation

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But, How Do We Stay Balanced?

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Need: three volunteers proud of their very diverse height.

Beautiful Balance (SIMPLEST version)

Insert(middle)

Insert(small)

Insert(tall)

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10But… BSTs are under-constrained in unfortunate ways;

ours may not look like this.

Bad Case #1(SIMPLEST version)

Insert(small)

Insert(middle)

Insert(tall)

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11How do we fix the bad case?

How do we transition among different possible trees?

So you say you want a revolution?

Scenario: k is one of the 1%; we want a revolution to depose k and give m root privileges.

What do we do?

Well, first…

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k

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m

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X

Well.. what do we know?

Scenario: k is one of the 1%; we want a revolution to depose k and give m root privileges.

What’s everything weknow about nodes k and m and subtrees X,Y, and Z?

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k

Z

m

Y

X

A real solution?

Scenario: k is one of the 1%; we want a revolution to depose k and give m root privileges.

Now, how can we makem the root?

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k

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m

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X

Single Rotation(SIMPLEST version)

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General Single Rotation

• After rotation, subtree’s height same as before insert!• Height of all ancestors unchanged.

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Zh h - 1

h + 1 h - 1

h + 2

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h - 1

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h - 1

h + 1

An insert made this BAD!

16Why couldn’t the bad insert be in X?Why does it matter that ancestors stay the same?

Bad Case #2(SIMPLEST version)

Insert(small)

Insert(tall)

Insert(middle)

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Double Rotation(SIMPLEST version)

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General Double Rotation

• Height of subtree still the same as it was before insert!• Height of all ancestors unchanged.

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X Y

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X Y

h

h - 1?

h - 1

h - 1

h + 2

h + 1

h - 1h - 1

h

h + 1

h

h - 1?

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Today’s Outline

• Addressing one of our problems• Single and Double Rotations• AVL Tree Implementation

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Insert Algorithm

• Find spot for value• Hang new node• Search back up for imbalance• If there is an imbalance:

case #1: Perform single rotation and exit

case #2: Perform double rotation and exit

Mirrored cases also possible

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Easy Insert

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Insert(3)

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Hard Insert (Bad Case #1)

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Insert(33)

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Single Rotation

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Hard Insert (Bad Case #2)

Insert(18)

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Single Rotation (oops!)

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Double Rotation (Step #1)

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Look familiar?18

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Double Rotation (Step #2)

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AVL Algorithm Revisited• Recursive1. Search downward for spot2. Insert node3. Unwind stack, correcting heights a. If imbalance #1, single rotate b. If imbalance #2, double rotate

• Iterative1. Search downward for spot, stacking parent nodes2. Insert node3. Unwind stack, correcting heights a. If imbalance #1, single rotate and exit b. If imbalance #2, double rotate and exit 29

Single Rotation Codevoid RotateLeft(Node *& root) { Node * temp = root->right; root->right = temp->left; temp->left = root; root->height = max(height(root->right), height(root->left)) + 1; temp->height = max(height(temp->right), height(temp->left) + 1; root = temp;}

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root

temp

(The “height” function returns -1 for a NULL subtreeor the “height” field of a non-NULL subtree.)

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Double Rotation Codevoid DoubleRotateLeft(Node *& root) { RotateRight(root->right); RotateLeft(root);}

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First Rotation

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Double Rotation Completed

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First Rotation Second Rotation

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AVL

• Automatically Virtually Leveled• Architecture for inVisible Leveling (the “in” is inVisible)

• All Very Low• Articulating Various Lines• Amortizing? Very Lousy!• Absolut Vodka Logarithms• Amazingly Vexing Letters

Adelson-Velskii Landis33

To Do

• Posted readings on priority queues and sorts, but de-emphasizing heaps

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Coming Up

• Quick Jump Back to Priority Queue ADT– A very familiar data structure gives us O(lg n)

performance!

• On to sorts• Then forward again!

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